Problem: What is the slope of the secant line that intersects the graph of $g(x)=3^{2x}$ at $x=0$ and $x=2$ ?
The secant line will pass through points $(0,g(0))$ and $(2,g(2))$. $\begin{aligned} \text{slope}&=\dfrac{\text{change in }y}{\text{change in }x} \\\\ &=\dfrac{g(2)-g(0)}{2-0} \end{aligned}$ ${0.5}$ ${1}$ ${1.5}$ ${2}$ ${\llap{-}0.5}$ ${10}$ ${20}$ ${30}$ ${40}$ ${50}$ ${60}$ ${70}$ ${80}$ $y$ $x$ $(0,g(0))$ $(2,g(2))$ secant line We will need to know the values of $g(0)$ and $g(2)$ to find the slope. $\begin{aligned} g(0)&=3^{2(0)} \\\\ &=1 \\\\ g(2)&=3^{2(2)} \\\\ &=81 \end{aligned}$ Now we can find the slope: $\begin{aligned} \dfrac{g(2)-g(0)}{2-0}&=\dfrac{81-1}{2} \\\\ &=40 \end{aligned}$ The slope of the secant line that intersects the graph of $g(x)=3^{2x}$ at $x=0$ and $x=2$ is $40$. Notice that the slope of the secant line is calculated just like the average rate of change over the interval.